Understanding Relations

Mathematical relation map 1 to 1 to the common definition of a relation.

You could, for example, say that 1<2 or 10=10; these relations have precise mathematically definitions.

We write: a<b If and only if a is less then b; that is, there exist a positive number n such that: a + n = b If there is any number we could add to a so that the product is equal to b; a has to be less then b.

Properties of relations

A relation can have certain named properties which are as follows:

• Reflexive: \forall a \in A : a \mathrel{\sim}_{R} a
• Symmetric: \forall a,b \in A : a \mathrel{\sim}_{R} b \Leftrightarrow b \mathrel{\sim}_{R} a
• Transitive: \forall a,b,c \in \mathbb{Z} : (a|b) \land (b|c) \Rightarrow (a|c)

Placeholder relations

You sometimes see placeholder relation like this: \mathrel{\sim}_R; where the \mathrel{\sim} is the placeholder and the _R indicates which kind of relation it is (the name). You can denote any kind of relation this way, but you need to give some context or outright explain which relation you want to represent.

Example: a \mathrel{\sim}_D b \Leftrightarrow \exists k \in \mathbb{Z} : b = k \cdot a

This is the formal definition of the | relations; which means that a can be divide b without a remainder. In other words b is a multiple of a.

Relations do not have to look like this, you are more likely to see "named" relations defined using all kind of symbols (for example: a \equiv_3 b \Leftrightarrow 3 | (a - b)).

Proving that | is transitive

We want to show that \forall a,b,c \in \mathbb{Z} : (a|b) \land (b|c) \Rightarrow (a|c).

Proof: Assume that that a,b,c \in \mathbb{Z}, and suppose that (a|b) and (b|c).

1. Since (a|b), there exists a n \in \mathbb{Z} such that: a \cdot n = b
2. Since (b|c) , there exists a m \in \mathbb{Z} such that: b \cdot m = c
3. If we substitute the expression for b into the equation for c: c = (a \cdot n) \cdot m = a \cdot (n \cdot m)
4. Since n \cdot m \in \mathbb{Z}, we conclude that (a|c); thus showing that the implication holds. \blacksquare